How to avoid being hit by a laser in a room of mirrors [video]

https://www.youtube.com/watch?v=kDZzyi-EaLQ

by sigil • 5 years ago

262 91 5 years ago

xwdv• 5 years ago

Incredible, but I don’t see a way to quickly figure out what the safe spot would be given any configuration of a room.

jekub• 5 years ago

It works the other way, the shooter position is fixed as well as your position. Your are allowed to place a finite amount of blockers to make your position safe. The video comment say that it is always possible to make your place safe with at most 16 blockers. There is a link to the actual code used.

makach• 5 years ago

Does this also work in a 3D environment?

mitko• 5 years ago

You could try projecting rays from start through the obstacles to see where they intersect after a limited number of reflections. This should narrow it to a few candidate spots

mehwoot• 5 years ago

There isn't always a safe spot, but you can generate a room that makes a safe spot- https://www.youtube.com/watch?v=jJ6FD59U0_E

laurent92• 5 years ago

The area covered by the lasers should remain painted, so we could see all the spots which haven’t been hit ever.

tdeck• 5 years ago

I'm curious why you couldn't just surround yourself with the 16 objects. Perhaps they're meant to be much smaller than the person?

ordu• 5 years ago

This way allow you to hide. A tight group of obstacles will give away your position and it will be demolished with a bazooka. Though an attacker might calculate the possible location of your body and use a grenade. But it might be unsafe for himself.

neitherboosh• 5 years ago

The posted solution seems to work even when the objects have zero area, and are just points in space. If you just surrounded the target with 16 points, there would either be gaps or the points would be infinitesimally close to the target and would occupy the same space.

Pyramus2• 5 years ago

Both person and obstacles are meant to be infinitesimally small, i.e. points with mass zero, which makes one really appreciate the beauty of the solution.

bunabhucan• 5 years ago

This video explains it. His pinned comment explains the mod arithmetic:

https://www.youtube.com/watch?v=jJ6FD59U0_E

parhamn4• 5 years ago

Thanks for the share! I took some quick (read, not proof-read) snippets on this if anyone wants to see: https://synth.app/s/uiElJvnnk

Pyramus2• 5 years ago

I wish he had explained the mod arithmetic more, the periodicity of 4 is really at the heart of the solution.

nullc• 5 years ago

If the room was an octagon would you need 64 objects?

seaish• 5 years ago

They did do it with a hexagon with joined sides. Not sure if the octagon is solvable. https://youtu.be/qhqdAiypldc

schappim• 5 years ago

Green Circle: You expect me to talk?

Red Circle: No, Mr Green Circle, I expect you to d̵i̵e̵ find the spot in the mirror room where you won't get hit!

Lio• 5 years ago

Surely it's more like:

[Red Circle]: I suggest a Duel of Titans, "mano a mano" the only true test for gentlemen.

[Valet leading Green Circle to room of mirrors]: If you kill him, all this be mine. Monsieur, Good shooting.

Although it does also rather make me think of Buster Keaton standing still as half a house falls over him and the cameraman closes his eyes.

Yeri• 5 years ago

This is stressful to watch.

irjustin• 5 years ago

Feels like the difference between playing the video game and watching TAS bot do the run =/

rcfaj7obqrkayhn• 5 years ago

for other people that also didn't get it... there's a green circle on the top right side corner

blu_• 5 years ago

What about scatter from the objects?

infogulch2• 5 years ago

The design assumes absorbing obstacles. He did make a separate video with reflecting obstacles, but the green circle is not safe.

https://youtube.com/shorts/DyRPASsynhI

wnkrshm• 5 years ago

Absorbing obstacles will technically not protect you either from receiving any light at all. Diffraction will 'push' some light into the geometric optical shadows.

mike_d• 5 years ago

Here is the explanation: https://www.youtube.com/watch?v=jJ6FD59U0_E

parhamn4• 5 years ago

I took some quick (read, not proof-read) snippets on this if anyone wants a quick run down: https://synth.app/s/uiElJvnnk

techdragon• 5 years ago

Thank you. So much more useful to grasp this relatively simple geometry problem with a few step by step pictures. Wish the link was to this instead of the YouTube clip.

fxtentacle• 5 years ago

That page is great, but the synth.app homepage seems weird to me.

"Browser from thefuture. Augment your work and your mind with the internet, don't just browse it."

What exactly is synth.app you offering? I couldn't figure it out from scrolling around their homepage a bit.

codetrotter2• 5 years ago

This is very neat. I feel strongly however, that each of the buttons that you have for opening the video ought to directly link to the time stamp in question.

So for the button you have for time stamp 02:46 you’d link to https://www.youtube.com/watch?v=jJ6FD59U0_E&t=2m46s and for the button for time stamp 03:30 you’d link to https://www.youtube.com/watch?v=jJ6FD59U0_E&t=3m30s and so on. For the very first button however, linking the very start of the video is nice though.

thunderbong• 5 years ago

This explains it perfectly!

antognini• 5 years ago

This method of images is also used to add reverberation to audio. If you have some audio and want to make it sound like it was spoken by someone in a room of a particular size, you do a similar thing where you tile the room and then draw lines from the source to every image of the microphone to calculate an impulse response function.

infogulch2• 5 years ago

Here's one with a close up of 'the circle with plot armor', as one comment describes:

https://youtu.be/jgpjnrFEUFI

bayesian_horse• 5 years ago

Don't go into the room.

shapefrog• 5 years ago

Break the mirrors.

taneq• 5 years ago

Calm down, Alexander! ;)

effingwewt• 5 years ago

Duck?

Does this always assume a 2-D room?

I wonder what the simulation would look like for a 3-D?

Makes me think of Catherine Zeta Jones in 'Entrapment', or Vincent Cassel in 'Ocean's 12'.

fouronnes3• 5 years ago

Ahah, your comment has the same vibe as this great fiction of how would Richard Feynmann pass a "lateral-thinking puzzle" interview question.

https://ericlippert.com/2011/02/14/what-would-feynman-do/

pvaldes• 5 years ago

Switch one receives a green cable, light C in the room has a green cable, Switch two has a yellow cable, same as light A in the room...

starfallg• 5 years ago

Turn off the power.

IncRnd• 5 years ago

The other day, when I was at a museum, I had to solve this particular problem. You see I really needed to get inside the room and move around after hours.

I spent a whole month making friends with the guards, only to have them change shift at the last minute. Thankfully, my cohort pointed out I could wear a reflective suit. All is well that ends well!

ARandomerDude• 5 years ago

Wear a full suit of armor made out of mirrors.

lvxferre• 5 years ago

Shoot the shooter before the shooter shoots you.

zzt1232• 5 years ago

This was beautiful. I never imagined the existence of a solution was always guaranteed, given that this includes the case where you are using point-particle blockers.

uuidgen• 5 years ago

I think the solution can be reached following from a set of simple observations about a single line of sight:

- in a square room all angles are right

- with wall at right angles the laser light will always create an inscribed square and return to the source for any given angle

- there is exactly one angle in a range <0,90> at which the light bounced from a single wall will go through an arbitrary point in the room

- there are 4 walls which gives at most 4 squares to be blocked

- the blocking point can be set anywhere on the inscribed square before the target - this gives at most 2 points per square (we can consider them left- and right- -hand directed) hence 8 points necessary

- at start we selected only angles from the <0,90> degree range for the ease of calculation, so there are also symmetric versions in the (90,180> range

- which gives us at most 16 points to block all the possible squares in both directions

zzt1232• 5 years ago

This line of reasoning reminds me of that one IMO geometry problem…

I’ll unwind this once I’m less brain foggy.

gampleman• 5 years ago

I wonder if there is a way to solve it such that both the shooter and the target are safe?

akdor11542• 5 years ago

It should be more like "how to find a shadow in a room full of mirrors", shouldn't it? Which is even more amazing.

lmilcin2• 5 years ago

Well... no.

These are two separate problems.

This example doesn't show all places where the shadow is.

It only shows, when you place yourself and a light, an example arrangement of occluders that will prevent you from being hit.

akdor11542• 5 years ago

Why are they separate problems? Maybe I should have phrased "how to create a complete shadow in a room full of mirrors".

SamBam2• 5 years ago

This example doesn't show all places where the shadow is.

I'm curious. Does it not? Because, from the descriptions of how to place all the blockers in order to create a shadow on that one spot, it did seem to me as if that one spot would be the only shadow in the room.

With 16 blockers can you have more that one discrete shadow?

NKosmatos3• 5 years ago

There was an interactive web page where you could place single point light sources and obstacles freely in a 2D room and then see in real time the shadows and where are the safe points. Too bad I can't remember the URL and can't find this old page. Anyone has a link to share?

NKosmatos3• 5 years ago

The closest I've found is this one, but it's not what I was referring to: https://ricktu288.github.io/ray-optics/

etherblood• 5 years ago

Maybe you mean this? https://www.redblobgames.com/articles/visibility/

azalemeth• 5 years ago

Fun problem. Can't help but ask what happens if the room is circular, or shaped in such a way that it doesn't easily tessellate, however...

toxik• 5 years ago

Interesting question, or even a simpler one: an isosceles triangle.

Cerium• 5 years ago

If the room was circular the laser beam would distort.

mnmmn1234562• 5 years ago

The solution is great, but it shows a fundamental flaw in mathematics, that I am thinking about a lot: You have to be smart, no doubt, but you also have to know the answer to get to the solution - It requires so many steps first until "the solution / the proof is clear". If you go down a different route (as I would have done), then you'd likely not succeed and certainly be in "rough" unknow territory, making it difficult for others to help or assess your approach. And this is expected: The hard problems will take the smartest people to solve eventually. I wish we could be more honest about that, that Math is a lot about "memorizing problems". And that was always the case (over history): At some point in history all this was a frontier. And there will be people willing to fight on this frontier and I am not disputing that intelligence/logical thinking is also required. I am just wondering if all this couldn't be put to more practical use. But then maybe it is and as fascinating as it is, most of us move on from Math and work on real world engineering tasks, eventually and from time to time still look back to these problems, as we do today. And absorb them, adding them to our repertoire and looking/assessing for real world application, which likely is limited / none.

antonfire2• 5 years ago

I think a bit of disillusionment with mathematics is healthy, but it's misguided to call this kind of thing "a fundamental flaw in mathematics".

For one thing, in my experience, working on real life problems involves a whole lot more "memorizing problems" than working on mathematics.

And thank goodness for that! When I drive over a bridge or install an app, I really don't want to hear that whoever made it has an aversion to memorizing problems and solutions. Solving a real world problem is usually boring. Making a real life thing well is usually about correctly putting together many pieces that some other people have put together, in a way that's roughly similar to how somebody else has already put a lot of those pieces together before. Most of the work is well-trodden and uncreative.

I think part of the reason mathematics feels like it's about "memorizing problems" is that getting something deeper than that out of it is a habit/skill. A very important aspect of the vague notion of "mathematical maturity" is a habit of looking at a solution to a problem with the mindset of "how could I have come up with this?". That is, unpacking a problem and solution into some deeper understanding or way of thinking that led to it. As opposed to filing the problem and solution away "as is" into some toolbox to be referenced later. A lot of "gotcha!" solutions to mathematical problems are unsatisfying precisely for this reason.

In a lot of cases once you've read a problem and read a solution to that problem, and understood the solution, you've done about 10% of the work. The remaining 90% is this difficult work of unpacking and repacking lessons from that solution into something that actually deepens your understanding of what the problem is about. Sometimes reading the solution is actually doing negative work.

In other words, if you look at this problem and look at this solution and think "neat, but I don't get anything out of it beyond 'neat'", that's normal and fine, and probably correct. But that doesn't mean there isn't anything in it beyond 'neat', and a big part of learning mathematics well is to dig deeper than that even when the problem and solution don't force you to. Whether that digging is worth it is up to you. (It's often kind of a crapshoot in terms of payoff.)

But that's a more complicated situation than "this shows a fundamental flaw in mathematics".

Traster• 5 years ago

It's certainly true that for a some mathematical problems memorizing the steps to the solution is often the most practical way forward (especially if you're planning on passing an exam), but there is definitely more to it. Most of the work proving things is actually exactly what you say - going down a different route. That's why mathematicians have careers, they spend days/months/years going down different routes and taking different appraoches to solve something. Naturally, they don't explain all the ways they didn't solve the problem when they finally do solve a problem. You don't need to know the answer to get to the solution. This is just like saying "The problem with a maze is you need to know the route through it before you can get through the maze"- no, what you do is you search for the route through the maze and once you've finished searching then you know the answer and suddenly you can walk through the maze repeatedly very easily.

And similarly, whilst one proof might have limited practical applications, it's very difficult to know which proofs do have practical applications and it's almost impossible to know which results might later yeild further proofs that do have practical applications.

None of this is really a matter of intelligence, it's a matter of putting in work (although obviously there is intelligence involved to find particularly neat ways of appraoching problems).

And why do mathematicians learn proofs? Often, to see the approaches that you can use to solve other problems.

antonfire2• 5 years ago

Naturally, they don't explain all the ways they didn't solve the problem when they finally do solve a problem.

I do think something related to this is less obvious than you're suggesting. A proof generally presents a train of thought gets you from point A to point B. It's almost never a representation of the train of thought that the author actually followed to get from point A to point B. This is obvious to people with experience in mathematics, but it's much less obvious before you have that experience. Even if you know it's true in principle, appreciating it viscerally and adapting the way you interact with mathematics to account for it is basically a lifelong task.

A big component of good mathematical communication is explaining what motivates a solution to a problem (or even what motivates the problems in the first place). Sure, you don't have to do that every time, but it's fair criticize a broad lack of this. A lot of mathematical communication is... not good in this sense. Notably, the problem/solution in the OP isn't really motivated by the author. A skilled reader will think about it themselves, but they need to learn that skill somewhere.

More or less on topic, here's an example where an experienced mathematician tries to reproduce one of Sylow's theorems without looking it up, and writes down a more-honest account of the thought process involved: https://gowers.wordpress.com/2011/12/10/group-actions-iv-int...

engineer_222• 5 years ago

I'm curious, what draws you to thinking about this problem so much lately?

mnmmn1234562• 5 years ago

Teachers/Professors always pretend that you can find it, eg. you get it as a homework. And it works that way, yes, kind of - as an excercise and for "relatively simple" challenges. But for tough problems it's a "memorization of the solution challenge" and not one where you lose too much time trying to figure them out. This particular one makes it pretty obvious as well: Even understanding the solution is a challenge.

dkersten2• 5 years ago

Encircle yourself with the circles? That would be finite and if you're thin enough, might even be minimal? Certainly easier to find the configuration than the actual solution.

(ok ok, it looks like the blockers are point-size, so this wouldn't work of couse)

tantalor2• 5 years ago

Not finite. Not even countably finite.

odd_perfect_num• 5 years ago

Or even countably infinite :) One could argue that all finite sets are countable since it's not too hard to find a surjection from the naturals to a given finite set.

(Intended to be mildly humorous.)

dkersten2• 5 years ago

I wrote the top of my comment before realising that the points were points and bot circles with a radius.

soapdog• 5 years ago

question: "How to avoid being hit by laser in a room of mirrors?"

me, a connoisseur: "This requires further reflections..."

GistNoesis• 5 years ago

So in an icosahedral room, let me guess : one would need 696729600 blockers ?

LightG• 5 years ago

Thanks, this is exactly what I needed at the right time.

anamexis• 5 years ago

Does anyone know if this is solvable in 3 dimensions (inside a mirrored cube)?

mabbo• 5 years ago

Based on the full explanation video, I suspect it probably is.

https://www.youtube.com/watch?v=jJ6FD59U0_E

But perhaps instead of 4^2 blockers, you'd need 4^3. But, I'm making guesses based on intuition, which makes for bad math.

raldi2• 5 years ago

I’d like to see a version of this where the barriers are colored and the lasers are monochrome, but when a laser hits the barrier, instead of disappearing, it takes on that color. That way, you could see which barrier saves the target at any point.

ghostbrainalpha• 5 years ago

That's a really cool idea.

raldi2• 5 years ago

The lasers could even leave a dim colored trail behind them, so that all the deadly space in the room gets shaded over time.

vmception• 5 years ago

I think this just helped me solve bullet hell games where it is tempting to evade so much when you can just stay in one part of the screen

danmaku!

j2kun• 5 years ago

Made a demo here: https://j2kun.github.io/assassin-puzzle/index.html

Blog post: https://jeremykun.com/2018/07/24/visualizing-an-assassin-puz...

Excellent video that I learned of this problem from: https://www.youtube.com/watch?v=a7gp9c2p0UQ

kazinator• 5 years ago

The chosen visualization complicates it. In a perfectly square room, a beam in any direction, originating from some interior point p0, will hit the four mirrors and form a rectangle, passing through the original p0 at the same angle. When tracing a beam, we don't have to consider more than five mirror strikes.

Moreover, this rectangle can only pass through another distinct point p1 exactly once. If we know that if p1 lies on a beam rectangle that is interrupted by an obstacle, there are only two possibilities: p1 lies on the unobstructed part of the rectangle between p0 and the obstruction. Or else it lies on the shaded part. We can determine which rectangles that pass through p0 also pass through p1, and then for each one determine whether p1 is in the shaded part.

I suspect this can be done with some computational geometry (with a fair number of cases in it) without resorting to a brute-force ray-casting technique.

EricRiese• 5 years ago

I thought this was going to be a new album by The Flaming Lips.

1_over_n• 5 years ago

How about just wrapping yourself in mirrors :)

How to avoid being hit by a laser in a room of mirrors [video]

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